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公司里有 n 名员工，每个员工的 ID 都是独一无二的，编号从 0 到 n - 1。公司的总负责人通过 headID 进行标识。


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        <h1 class="title">leetcode5354. 通知所有员工所需的时间</h1>
        <div class="stuff">
            <span>三月 14, 2020</span>
            

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            <h1 id="leetcode5354-通知所有员工所需的时间"><a href="#leetcode5354-通知所有员工所需的时间" class="headerlink" title="leetcode5354. 通知所有员工所需的时间"></a>leetcode5354. 通知所有员工所需的时间</h1><hr>
<blockquote>
<p>公司里有 n 名员工，每个员工的 ID 都是独一无二的，编号从 0 到 n - 1。公司的总负责人通过 headID 进行标识。</p>
</blockquote>
<blockquote>
<p>在 manager 数组中，每个员工都有一个直属负责人，其中 manager[i] 是第 i 名员工的直属负责人。对于总负责人，manager[headID] = -1。题目保证从属关系可以用树结构显示。</p>
</blockquote>
<blockquote>
<p>公司总负责人想要向公司所有员工通告一条紧急消息。他将会首先通知他的直属下属们，然后由这些下属通知他们的下属，直到所有的员工都得知这条紧急消息。</p>
</blockquote>
<blockquote>
<p>第 i 名员工需要 informTime[i] 分钟来通知它的所有直属下属（也就是说在 informTime[i] 分钟后，他的所有直属下属都可以开始传播这一消息）。</p>
</blockquote>
<blockquote>
<p>返回通知所有员工这一紧急消息所需要的 分钟数 。</p>
</blockquote>
<blockquote>
<p>示例 1：<br>输入：n = 1, headID = 0, manager = [-1], informTime = [0]<br>输出：0<br>解释：公司总负责人是该公司的唯一一名员工。</p>
</blockquote>
<blockquote>
<p>示例 2：<br><img src="https://imgconvert.csdnimg.cn/aHR0cHM6Ly9hc3NldHMubGVldGNvZGUtY24uY29tL2FsaXl1bi1sYy11cGxvYWQvdXBsb2Fkcy8yMDIwLzAzLzA4L2dyYXBoLnBuZw?x-oss-process=image/format,png" alt="在这里插入图片描述"><br>输入：n = 6, headID = 2, manager = [2,2,-1,2,2,2], informTime = [0,0,1,0,0,0]<br>输出：1<br>解释：id = 2 的员工是公司的总负责人，也是其他所有员工的直属负责人，他需要 1 分钟来通知所有员工。<br>上图显示了公司员工的树结构。</p>
</blockquote>
<p>这场周赛前两题算出来了，到这道题就卡住了。<br>思路：用一个map存储上级和他小弟之间的对应关系，之后像层序遍历一样遍历key（上级）。</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">numOfMinutes</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> headID, <span class="keyword">int</span>[] manager, <span class="keyword">int</span>[] informTime)</span> </span>&#123;</span><br><span class="line">       HashMap&lt;Integer,ArrayList&lt;Integer&gt;&gt;map=<span class="keyword">new</span> HashMap&lt;&gt;();</span><br><span class="line">       <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; manager.length; i++) &#123;</span><br><span class="line">       <span class="comment">//map的初始化</span></span><br><span class="line">           <span class="keyword">if</span>(!map.containsKey(manager[i])) &#123;</span><br><span class="line">               map.put(manager[i], <span class="keyword">new</span> ArrayList&lt;&gt;());</span><br><span class="line">           &#125;</span><br><span class="line">           map.get(manager[i]).add(i);</span><br><span class="line">           <span class="comment">//list里面存的是编号ID</span></span><br><span class="line">       &#125;</span><br><span class="line">       <span class="keyword">int</span>[]time=<span class="keyword">new</span> <span class="keyword">int</span>[n];<span class="comment">//每个人被通知到需要多少时间</span></span><br><span class="line">       Queue&lt;Integer&gt;queue=<span class="keyword">new</span> LinkedList&lt;&gt;();</span><br><span class="line">       queue.offer(headID);</span><br><span class="line">       <span class="keyword">while</span>(!queue.isEmpty())&#123;</span><br><span class="line">           <span class="keyword">int</span> boss=queue.remove();</span><br><span class="line">           ArrayList&lt;Integer&gt;list=map.get(boss);</span><br><span class="line">           <span class="keyword">if</span>(list!=<span class="keyword">null</span>)&#123;</span><br><span class="line">               <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;list.size();i++)&#123;</span><br><span class="line">               <span class="comment">//当前这个人的时间=上级被通知的时间+上级通知我的时间</span></span><br><span class="line">               time[list.get(i)]=time[boss]+informTime[boss];</span><br><span class="line">               queue.offer(list.get(i));</span><br><span class="line">           &#125;</span><br><span class="line">           &#125;</span><br><span class="line">           </span><br><span class="line">       &#125;</span><br><span class="line">       <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line">       <span class="comment">//找最后一个被通知的人的时间</span></span><br><span class="line">       <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;time.length;i++)</span><br><span class="line">           ans=Math.max(ans,time[i]);</span><br><span class="line">       <span class="keyword">return</span> ans;</span><br><span class="line">   &#125;</span><br></pre></td></tr></table></figure>
<p><strong>leetcode 51/100</strong></p>

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